Respuesta :
Explanation:
Given that,
The slope of the ramp, [tex]\theta=8^{\circ}[/tex]
Mass of the box, m = 60 kg
(a) Distance covered by the truck up the slope, d = 300 m
Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :
[tex]W=\dfrac{1}{2}m(v^2-u^2)=0[/tex]
u and v are the initial and the final velocity of the truck
(b) The work done on the box by the force of gravity is given by :
[tex]W=Fd\ cos\theta[/tex]
Here, [tex]\theta=90+8=98^{\circ}[/tex]
[tex]W=mgd\ cos\theta[/tex]
[tex]W=60\times 9.8\times 300\ cos(98)[/tex]
W = -24550.13 J
(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.
(d) The work done by friction is given by :
[tex]W_f=-W[/tex]
[tex]W_f=24550.13\ J[/tex]
Hence, this is the required solution.
